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# P2 Question watch

1. Ok I have the curve
Y=5cosx-2sinx
Now I found all the stuff like R and alpha when it's equivalent to Rcos(x+alpha).
Then it asks me to find the intercept on the Y axis. Done that. But then it goes on to ask me about finding the minimum point on the curve. I have no idea how to do this. Any help is appreciated.

(It's Q3 on the Solomon Paper B).
2. Bump...
3. (Original post by Les Paul)
Ok I have the curve
Y=5cosx-2sinx
Now I found all the stuff like R and alpha when it's equivalent to Rcos(x+alpha).
Then it asks me to find the intercept on the Y axis. Done that. But then it goes on to ask me about finding the minimum point on the curve. I have no idea how to do this. Any help is appreciated.

(It's Q3 on the Solomon Paper B).
Differentiate?
4. (Original post by Les Paul)
Ok I have the curve
Y=5cosx-2sinx
Now I found all the stuff like R and alpha when it's equivalent to Rcos(x+alpha).
Then it asks me to find the intercept on the Y axis. Done that. But then it goes on to ask me about finding the minimum point on the curve. I have no idea how to do this. Any help is appreciated.

(It's Q3 on the Solomon Paper B).
The minimum is going to be when y = -R, since the minimum value for cos(anything) is always -1. You can find out the values of x which this applies to by knowing your value of alpha and thinking about what values of x + alpha would cause cos(x + alpha) to equal -1.

Ben
5. (Original post by Les Paul)
Ok I have the curve
Y=5cosx-2sinx
Now I found all the stuff like R and alpha when it's equivalent to Rcos(x+alpha).
Then it asks me to find the intercept on the Y axis. Done that. But then it goes on to ask me about finding the minimum point on the curve. I have no idea how to do this. Any help is appreciated.

(It's Q3 on the Solomon Paper B).
I'd probably expand the Rcos(x+a) term to Rcosxcosa - Rsinxsina and then equate that to y = 5cosx - 2sinx.

Therefore you should get Rcosa = 5 and Rsina = 2 (a standing for alpha).
Divide sina by cosa to get get tana = 2/5 and use pythagoras to find the value of R (it should be the square root of 2^2 + 5^2). Therefore alpha = 21.8 degrees and R is root 29. i.e. 5cosx - 2sinx can be written as 5.39cos(x+21.8). You know the maximum and minimum will thus be 5.39 and -5.39 (both are just root +/- 29), respectively, and the points of these values can also be determined by a quick sketch.
6. (Original post by mik1a)
Differentiate?
Not for P2.
7. (Original post by boygenious)
I'd probably expand the Rcos(x+a) term to Rcosxcosa - Rsinxsina and then equate that to y = 5cosx - 2sinx.

Therefore you should get Rcosa = 5 and Rsina = 2 (a standing for alpha).
Divide sina by cosa to get get tana = 2/5 and use pythagoras to find the value of R (it should be the square root of 2^2 + 5^2). Therefore alpha = 21.8 degrees and R is root 29. i.e. 5cosx - 2sinx can be written as 5.39cos(x+21.8). You know the maximum and minimum will thus be 5.39 and -5.39 (both are just root +/- 29), respectively, and the points of these values can also be determined by a quick sketch.
Oh ok. Thanks a lot man.

yeah differentiation of Trigonometric equations aren't included in P2.

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